Fourier series. Complex Fourier series. Positive and negative frequencies. Fourier series demonstration. Notes. Notes. Notes.

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1 Fourier series Fourier series of a periodic function f (t) with period T and corresponding angular frequency ω /T : f (t) a 0 + (a n cos(nωt) + b n sin(nωt)), n1 Fourier series is a linear sum of cosine and sine functions with discrete frequencies that are integer multiples of the frequency of f (t). This gives rise to a discrete frequency spectrum given by the Fourier coefficients (frequency amplitudes). Complex Fourier series Complex representation of Fourier series of a function f (t) with period T and corresponding angular frequency ω /T : f (t) n c n e inωt, where (a n ib n )/, n > 0, c n a 0 / n 0, (a n + ib n )/ n < 0 Note that the summation goes from to. Now have a negative frequencies as well. Positive and negative frequencies Positive frequencies (n > 0): e inωt cos(nωt) + i sin(nωt) Negative frequencies (n < 0): e i n ωt cos( n ωt) i sin( n ωt) Fourier series demonstration Applet for Fourier Series:

2 The Fourier transform For non-periodic (or periodic) functions f (t), we can define the Fourier transform Fourier transform F[f (t)] g(ω) 1 inverse transform F 1 [g(ω)] f (t) 1 g(ω)e iωt dω, where g(ω) corresponds to a continuous frequency spectrum. Properties of the Fourier transform Linearity F[f (t) + g(t)] F[f (t)] + F[g(t)] Shifting F[f (t t 0 )] e iωt0 g(ω) F 1 [g(ω ω 0 )] e iω0t f (t) Scaling F[f (αt)] 1 α g(ω/α) F 1 [g(βt)] 1 β f (t/β) Power spectrum F[f (t)] (F[f (t)]) F[f (t t 0 )] g(ω) cannot reconstruct f(t) from power spectrum since phase information is lost. Note: F[f (t t 0 )] F[f (t)] Parseval s theorem f (t) dt }{{} Power δ-function δ(x) 1 g(ω) }{{} Power spectrum eixt dt δ( x) δ(x) dω Fourier transform of a derivative Fourier transform can be used to solve differential equations. Consider F[f (t)] 1 + f (t)e iωt dt (1) where f (t) df /dt. Integrating by parts, we obtain F[f (t)] e iωt f (t) + iω + () Since f (t) vanishes as t ±, the first term vanishes and we have F[f (t)] iωf[f (t)] (3) Can show that for n th derivative F[f n (t)] (iω) n F[f (t)] The Fourier transform obeys certain symmetries. Consider the Fourier transform and its complex conjugate: g(ω) g (ω) f (t)e iωt dt If f (t) is real (f (t) f (t)), then g (ω) 1 f (t)e i( ω)t dt g( ω)

3 g (ω) g( ω) means the real part of the transform Re(g(ω)) is even, while Im(g(ω)) is odd: g(ω) Re(g(ω)) + iim(g(ω)) g( ω) Re(g( ω)) + iim(g( ω)) g (ω) Re(g(ω)) iim(g(ω)) Conversely, if f (t) is purely imaginary, then g (ω) 1 f (t)e i( ω)t dt g( ω) Hence, g( ω) g (ω), which means that Re(g(ω)) is odd, while Im(g(ω)) is even. Let f (t) be even: f ( t) f (t), then can use scaling property to show: But, F[f ( t)] F[f (( 1)t)] 1 1 g(ω/( 1)) g( ω) F[f ( t)] F[f (t)] g(ω) Therefore, g(ω) g( ω). The transform is even too. Similarly, can show that if f (t) is odd, then g(ω) is odd as well. The symmetry properties of the Fourier transform can be summarized as follows: f (t) real f (t) imaginary f (t) even f (t) odd f (t) real and even f (t) real and odd f (t) imaginary and even f (t) imaginary and odd Re(g(ω)) even and Im(g(ω)) odd Re(g(ω)) odd and Im(g(ω)) even g(ω) even g(ω) odd g(ω) real and even g(ω) imaginary and odd g(ω) imaginary and even g(ω) real and odd Fourier Series vs. Fourier transform Let s compare Fourier series, complex representation of Fourier series and Fourier transform: Consider f (t) sin(ωt), where ω /T. What are its real Fourier coefficients? f (t) a 0 + (a n cos(nωt) + b n sin(nωt)), (4) n1 By inspection, a n 0, b 1 1. b n 0 for n 1.

4 Now let s find the Fourier coeffcients of f (t) sin(ωt) for the complex representation of the Fourier series: f (t) c n e inωt, (5) where, c n 1 T Substituting f (t), we obtain c n 1 T 1 T T / T / T / T / n T / T / sin(ωt)(cos(nωt) i sin(nωt))dt sin(ωt) cos(nωt)dt } {{ } 0 f (t)e inωt dt (6) T / i T / sin(ωt) sin(nωt))dt π c n i sin(x) sin(nx)) dx π }{{} πδ 1n { i n 1 + i n 1 Hence, c 1 i/ and c 1 i/, all other c n are zero. Note: A single harmonic (sin or cos) is represented by two Fourier coefficients in the complex Fourier series. Now consider the Fourier transform of f (t) sin(ωt). Rewrite as f (t) eiωt e iωt i : g(ω) e iωt e iωt e iωt dt i i 1 e i(ω ω)t dt 1 e i(ω+ω)t dt }{{}}{{} δ(ω ω) δ( (Ω+ω))δ(Ω+ω) i (δ(ω + ω) δ(ω ω)) Note that δ(x) δ( x). Therefore, F[sin(Ωt)] i (δ(ω + ω) δ(ω ω)) yields two δ-function peaks at ω ±Ω with imaginary amplitudes. This is analogous to the two complex Fourier coefficients we obtained earlier. Similarly, one can obtain F[cos(Ωt)] (δ(ω + ω) + δ(ω ω))

5 What is the Fourier transform of a constant function f (t) C? g(ω) C 1 C δ(ω) Ce iωt dt e iωt dt } {{ } δ( ω)δ(ω) δ-function peak at ω 0. Zero frequency mode. Fourier transform of periodic functions Fourier transform can operate on non-periodic, but also periodic functions, which we can express in terms of Fourier series. Let f (t) be a periodic function with angular frequency Ω: f (t) a 0 + (a n cos(nωt) + b n sin(nωt)), n1 Therefore, [ a ] 0 F[f (t)] F + (a n cos(nωt) + b n sin(nωt)), n1 Now use linearity of transform F[f (t)] F[a 0 /] + (a n F[cos(nΩt)] + b n F[sin(nΩt)]) n1 Fourier transform of periodic functions F[f (t)] a 0 δ(ω) + a n (δ(nω + ω) + δ(nω ω)) + i n1 b n (δ(nω + ω) δ(nω ω)) n1 Therefore, Fourier transforms of periodic functions yield a sum of δ-functions located at integer multiples of the frequency Ω. Gaussian peak: f (t) e αt F[f (t)] 1 e ω 4α 1 α e ω 4α e αt e iωt dt ( e α t + iω iω t+( α α) ( α) iω ) dt iω α(t+ e α ) iω e α( α) dt e αx dx }{{} π/α

6 Table of common Fourier transforms f (t) g(ω) 1 δ(t) constant C C δ(ω) sin(ωt) i δ(ω+ω) δ(ω Ω) cos(ωt) δ(ω+ω)+δ(ω Ω) Gaussian (α > 0): e αt 1 α e ω /(4α) Exponential (α > 0): e α t α π α +ω